Let $f : [ - 1,2 ] \rightarrow [ 0 , \infty )$ be a continuous function such that $f ( x ) = f ( 1 - x )$ for all $x \in [ - 1,2 ]$. Let $R _ { 1 } = \int _ { - 1 } ^ { 2 } x f ( x ) d x$, and $R _ { 2 }$ be the area of the region bounded by $y = f ( x ) , x = - 1 , x = 2$, and the $x$-axis. Then
(A) $R _ { 1 } = 2 R _ { 2 }$
(B) $R _ { 1 } = 3 R _ { 2 }$
(C) $2 R _ { 1 } = R _ { 2 }$
(D) $3 R _ { 1 } = R _ { 2 }$

For a point $P$ in the plane, let $d_1(P)$ and $d_2(P)$ be the distances of the point $P$ from the lines $x - y = 0$ and $x + y = 0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2 \leq d_1(P) + d_2(P) \leq 4$, is

Area of the region $\left\{ ( x , y ) \in \mathbb { R } ^ { 2 } : y \geq \sqrt { | x + 3 | } , 5 y \leq x + 9 \leq 15 \right\}$ is equal to
(A) $\frac { 1 } { 6 }$
(B) $\frac { 4 } { 3 }$
(C) $\frac { 3 } { 2 }$
(D) $\frac { 5 } { 3 }$

If the line $x = \alpha$ divides the area of region $R = \left\{ ( x , y ) \in \mathbb { R } ^ { 2 } : x ^ { 3 } \leq y \leq x , 0 \leq x \leq 1 \right\}$ into two equal parts, then
[A] $0 < \alpha \leq \frac { 1 } { 2 }$
[B] $\frac { 1 } { 2 } < \alpha < 1$
[C] $2 \alpha ^ { 4 } - 4 \alpha ^ { 2 } + 1 = 0$
[D] $\alpha ^ { 4 } + 4 \alpha ^ { 2 } - 1 = 0$

A farmer $F _ { 1 }$ has a land in the shape of a triangle with vertices at $P ( 0,0 ) , Q ( 1,1 )$ and $R ( 2,0 )$. From this land, a neighbouring farmer $F _ { 2 }$ takes away the region which lies between the side $P Q$ and a curve of the form $y = x ^ { n } ( n > 1 )$. If the area of the region taken away by the farmer $F _ { 2 }$ is exactly $30 \%$ of the area of $\triangle P Q R$, then the value of $n$ is $\_\_\_\_$.

The area of the region $\left\{ ( x , y ) : x y \leq 8,1 \leq y \leq x ^ { 2 } \right\}$ is
(A) $16 \log _ { e } 2 - \frac { 14 } { 3 }$
(B) $8 \log _ { e } 2 - \frac { 14 } { 3 }$
(C) $16 \log _ { e } 2 - 6$
(D) $8 \log _ { e } 2 - \frac { 7 } { 3 }$

Let the functions $f : \mathbb { R } \rightarrow \mathbb { R }$ and $g : \mathbb { R } \rightarrow \mathbb { R }$ be defined by
$$f ( x ) = e ^ { x - 1 } - e ^ { - | x - 1 | } \quad \text { and } \quad g ( x ) = \frac { 1 } { 2 } \left( e ^ { x - 1 } + e ^ { 1 - x } \right)$$
Then the area of the region in the first quadrant bounded by the curves $y = f ( x ) , y = g ( x )$ and $x = 0$ is
(A) $( 2 - \sqrt { 3 } ) + \frac { 1 } { 2 } \left( e - e ^ { - 1 } \right)$
(B) $( 2 + \sqrt { 3 } ) + \frac { 1 } { 2 } \left( e - e ^ { - 1 } \right)$
(C) $( 2 - \sqrt { 3 } ) + \frac { 1 } { 2 } \left( e + e ^ { - 1 } \right)$
(D) $( 2 + \sqrt { 3 } ) + \frac { 1 } { 2 } \left( e + e ^ { - 1 } \right)$

Let $f : [ 0,1 ] \rightarrow [ 0,1 ]$ be the function defined by $f ( x ) = \frac { x ^ { 3 } } { 3 } - x ^ { 2 } + \frac { 5 } { 9 } x + \frac { 17 } { 36 }$. Consider the square region $S = [ 0,1 ] \times [ 0,1 ]$. Let $G = \{ ( x , y ) \in S : y > f ( x ) \}$ be called the green region and $R = \{ ( x , y ) \in S : y < f ( x ) \}$ be called the red region. Let $L _ { h } = \{ ( x , h ) \in S : x \in [ 0,1 ] \}$ be the horizontal line drawn at a height $h \in [ 0,1 ]$. Then which of the following statements is(are) true?
(A) There exists an $h \in \left[ \frac { 1 } { 4 } , \frac { 2 } { 3 } \right]$ such that the area of the green region above the line $L _ { h }$ equals the area of the green region below the line $L _ { h }$
(B) There exists an $h \in \left[ \frac { 1 } { 4 } , \frac { 2 } { 3 } \right]$ such that the area of the red region above the line $L _ { h }$ equals the area of the red region below the line $L _ { h }$
(C) There exists an $h \in \left[ \frac { 1 } { 4 } , \frac { 2 } { 3 } \right]$ such that the area of the green region above the line $L _ { h }$ equals the area of the red region below the line $L _ { h }$
(D) There exists an $h \in \left[ \frac { 1 } { 4 } , \frac { 2 } { 3 } \right]$ such that the area of the red region above the line $L _ { h }$ equals the area of the green region below the line $L _ { h }$

Let $n \geq 2$ be a natural number and $f : [ 0,1 ] \rightarrow \mathbb { R }$ be the function defined by
$$f ( x ) = \begin{cases} n ( 1 - 2 n x ) & \text { if } 0 \leq x \leq \frac { 1 } { 2 n } \\ 2 n ( 2 n x - 1 ) & \text { if } \frac { 1 } { 2 n } \leq x \leq \frac { 3 } { 4 n } \\ 4 n ( 1 - n x ) & \text { if } \frac { 3 } { 4 n } \leq x \leq \frac { 1 } { n } \\ \frac { n } { n - 1 } ( n x - 1 ) & \text { if } \frac { 1 } { n } \leq x \leq 1 \end{cases}$$
If $n$ is such that the area of the region bounded by the curves $x = 0 , x = 1 , y = 0$ and $y = f ( x )$ is 4, then the maximum value of the function $f$ is

Let $\mathbb { R }$ denote the set of all real numbers. Then the area of the region
$$\left\{ ( x , y ) \in \mathbb { R } \times \mathbb { R } : x > 0 , y > \frac { 1 } { x } , 5 x - 4 y - 1 > 0,4 x + 4 y - 17 < 0 \right\}$$
is

(A)	$\frac { 17 } { 16 } - \log _ { e } 4$	(B)	$\frac { 33 } { 8 } - \log _ { e } 4$
(C)	$\frac { 57 } { 8 } - \log _ { e } 4$	(D)	$\frac { 17 } { 2 } - \log _ { e } 4$

The area enclosed between the curves $y ^ { 2 } = x$ and $y = | x |$ is
(1) $2 / 3$
(2) 1
(3) $1 / 6$
(4) $1 / 3$

The area (in square units) bounded by the curves $y = \sqrt{x}$, $2y - x + 3 = 0$, $X$-axis and lying in the first quadrant is
(1) 18 sq. units
(2) $\frac{27}{4}$ sq. units
(3) 9 sq. units
(4) 36 sq. units

The area (in sq. unit) of the region described by $A = \left\{ ( x , y ) : x ^ { 2 } + y ^ { 2 } \leq 1 \text{ and } y ^ { 2 } \leq 1 - x \right\}$ is
(1) $\frac { \pi } { 2 } - \frac { 2 } { 3 }$
(2) $\frac { \pi } { 2 } + \frac { 2 } { 3 }$
(3) $\frac { \pi } { 2 } + \frac { 4 } { 3 }$
(4) $\frac { \pi } { 2 } - \frac { 4 } { 3 }$

Let $A = \left\{ ( x , y ) : y ^ { 2 } \leq 4 x , y - 2 x \geq - 4 \right\}$. The area of the region $A$ in square units is
(1) 10
(2) 8
(3) 9
(4) 11

The area of the region (in square units) above the $x$-axis bounded by the curve $y = \tan x , 0 \leq x \leq \frac { \pi } { 2 }$ and the tangent to the curve at $x = \frac { \pi } { 4 }$ is
(1) $\frac { 1 } { 2 } \left( \log 2 - \frac { 1 } { 2 } \right)$
(2) $\frac { 1 } { 2 } ( 1 + \log 2 )$
(3) $\frac { 1 } { 2 } ( 1 - \log 2 )$
(4) $\frac { 1 } { 2 } \left( \log 2 + \frac { 1 } { 2 } \right)$

The area (in sq. units) of the region described by $\left\{ ( x , y ) : y ^ { 2 } \leq 2 x \text{ and } y \geq 4 x - 1 \right\}$ is
(1) $\frac { 9 } { 32 }$ sq. units
(2) $\frac { 7 } { 32 }$ sq. units
(3) $\frac { 5 } { 64 }$ sq. units
(4) $\frac { 15 } { 64 }$ sq. units

The area (in sq. units) of the region $\{(x, y) : x \geq 0,\ x + y \leq 3,\ x^2 \leq 4y$ and $y \leq 1 + \sqrt{x}\}$ is
(1) $\dfrac{59}{12}$ sq. units
(2) $\dfrac{3}{2}$ sq. units
(3) $\dfrac{7}{3}$ sq. units
(4) $\dfrac{5}{2}$ sq. units

The area (in sq. units) of the smaller portion enclosed between the curves, $x ^ { 2 } + y ^ { 2 } = 4$ and $y ^ { 2 } = 3 x$, is:
(1) $\frac { 1 } { \sqrt { 3 } } + \frac { 4 \pi } { 3 }$
(2) $\frac { 1 } { \sqrt { 3 } } + \frac { 2 \pi } { 3 }$
(3) $\frac { 1 } { 2 \sqrt { 3 } } + \frac { \pi } { 3 }$
(4) $\frac { 1 } { 2 \sqrt { 3 } } + \frac { 2 \pi } { 3 }$

Let $g ( x ) = \cos x ^ { 2 } , f ( x ) = \sqrt { x }$, and $\alpha , \beta ( \alpha < \beta )$ be the roots of the quadratic equation $18 x ^ { 2 } - 9 \pi x + \pi ^ { 2 } = 0$. Then the area (in sq. units) bounded by the curve $y = ( g o f ) ( x )$ and the lines $x = \alpha , x = \beta$ and $y = 0$, is
(1) $\frac { 1 } { 2 } ( \sqrt { 2 } - 1 )$
(2) $\frac { 1 } { 2 } ( \sqrt { 3 } - 1 )$
(3) $\frac { 1 } { 2 } ( \sqrt { 3 } + 1 )$
(4) $\frac { 1 } { 2 } ( \sqrt { 3 } - \sqrt { 2 } )$

The area (in sq. units) of the region $\{ x \in R : x \geq 0 , y \geq 0 , y \geq x - 2$ and $y \leq \sqrt { x } \}$ is
(1) $\frac { 13 } { 3 }$
(2) $\frac { 8 } { 3 }$
(3) $\frac { 10 } { 3 }$
(4) $\frac { 5 } { 3 }$

The area (in sq. units) of the region $\{ x \in R : x \geq 0 , y \geq 0 , y \geq x - 2$ and $y \leq \sqrt { x } \}$, is
(1) $\frac { 13 } { 3 }$
(2) $\frac { 10 } { 3 }$
(3) $\frac { 5 } { 3 }$
(4) $\frac { 8 } { 3 }$

Let $A(4,-4)$ and $B(9,6)$ be points on the parabola, $y^2 = 4x$. Let $C$ be chosen on the arc $AOB$ of the parabola, where $O$ is the origin, such that the area of $\triangle ACB$ is maximum. Then, the area (in sq. units) of $\triangle ACB$, is:
(1) 32
(2) $31\frac{3}{4}$
(3) $30\frac{1}{2}$
(4) $31\frac{1}{4}$

If the area enclosed between the curves $y = k x ^ { 2 }$ and $x = k y ^ { 2 } , ( k > 0 )$, is 1 sq. unit. Then $k$ is
(1) $\sqrt { 3 }$
(2) $\frac { 1 } { \sqrt { 3 } }$
(3) $\frac { \sqrt { 3 } } { 2 }$
(4) $\frac { 2 } { \sqrt { 3 } }$

The area (in sq. units) of the region $A = \{(x,y) \in R \times R \mid 0 \leq x \leq 3, 0 \leq y \leq 4, y \leq x^2 + 3x\}$ is
(1) $\frac{26}{3}$
(2) $8$
(3) $\frac{53}{6}$
(4) $\frac{59}{6}$

The area (in sq. units) of the region $A = \left\{ ( x , y ) : x ^ { 2 } \leq y \leq x + 2 \right\}$ is
(1) $\frac { 13 } { 6 }$
(2) $\frac { 31 } { 6 }$
(3) $\frac { 9 } { 2 }$
(4) $\frac { 10 } { 3 }$